In triangle ABC, A is 45 degrees, C is 30 degrees. Height AD = 30cm. Find the sides of the triangle.

Given:

ABC – triangle;

Angle A = 45 degrees;

Angle C = 30 degrees;

Height AD = 30 cm;

Find the sides of the triangle AB, AC, BC.

Decision:

1) Find angle B.

Since the sum of the angles of the triangle is 180 degrees, then we get:

Angle A + angle B + angle C = 180;

45 + angle C + 30 = 180;

Angle C = 180 – 45 – 30 = 150 – 45 = 105.

2) AD = AC * sin c;

Hence, AC = AD / sin C = 30 / sin 30 = 30 / (1/2) = 30 * 2 = 60 cm;

3) The theorem of sines.

BC / sin a = AC / sin b = AB / sin c;

BC / sin a = AC / sin b;

BC / sin 45 = 60 / sin 30;

BC = 60 * sin 45 / sin 30 = 60 / (1/2) * sin 45 = 120 * √2 / 2 = 60√2;

4) Cosine theorem:

AB ^ 2 = AC ^ 2 + BC ^ 2 – 2 * AC * BC * cos A = 3600 + 3600 * 2 – 2 * 60 * 60√2 * √ (1 – sin ^ 2 a) = 3 * 3600 – 7200 √2 * √2 / 2 = 3 * 3600 – 7200 = 3600;

AB = 60 cm.

Answer: AB = AC = 60 cm, BC = 60√2.